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I found in a book that if $H$ is the mean curvature vector field of the sphere $S^n(r)$ in $\mathbb{R}^{n+1}$, then $||H||^2 = \frac{1}{r^2}$ in every point of $S^n(r)$. I know that, if $x\in S^n(r)$ and $e_1, \ldots, e_n$ is an ortonormal basis of $T_x S^n(r)$, then $H(x) = \frac{1}{n}\sum_{i=1}^n h(e_i,e_i)$, where $h$ is the second fundamental form of the sphere in $\mathbb{R}^{n+1}$, but I am stuck here.

How should I prove that $||H||^2 = \frac{1}{r^2}$?

cos_dm_math21
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    Have you computed $h$? That's where I would start. (Also, last time I checked the mean curvature of the sphere of radius $r$ is $r^{-1}$, not $r^{-2}$.) –  Jan 22 '21 at 15:35
  • @PeterMorfe, you are right, it was $||H||^2 = \frac{1}{r^2}$. I fix it! Thank you. – cos_dm_math21 Jan 22 '21 at 16:44

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Let $\alpha$ be a unit speed curve on the sphere. Then to find $h_{\alpha(t)}(\alpha'(t),\alpha'(t))$, we project $\alpha''(t)$ in the normal direction of the sphere. So $$h_{\alpha(t)}(\alpha'(t),\alpha'(t))=\frac{\langle \alpha''(t),\alpha(t)\rangle}{\langle \alpha(t),\alpha(t)\rangle}\alpha(t)= -\frac{1}{r^2}\alpha(t),$$since $\langle \alpha'(t),\alpha(t)\rangle=0$ leads to $\langle \alpha''(t),\alpha(t)\rangle = -1$ by differentiating. If $N(p)=p/r$ is the outward unit normal field along the sphere, we have shown that $$h_p(v,v)=-\frac{1}{r}N(p)$$for all unit tangent vectors $v$. Let $v$ range over an orthonormal basis of the tangent space and average $n$ copies of this to get $$H(p)=\frac{1}{n}\left(-\frac{n}{r}N(p)\right)= -\frac{1}{r}N(p)\implies \|H(p)\|=\frac{1}{r}.$$Also, adjusting the scale and polarizing, we see that $$h_p(v,w)=-\frac{1}{r}\langle v,w\rangle N(p)$$for all tangent vectors $v$ and $w$ at $p$ (no longer restricted to be unit).

Ivo Terek
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  • I understand everything, unless why $h_{\alpha(t)}(\alpha'(t), \alpha'(t))$ is the projection of $\alpha''(t)$ on the normal direction of the sphere in the point $\alpha(t)$. By the definition I know, $h(X,Y) = \widetilde{\nabla}_X Y - \nabla _X Y$, where $\widetilde{\nabla}$ is the Levi-Civita connection of $\mathbb{R}^{n+1}$ and $\nabla$ is the Levi-Civita connection of the sphere. Did you use that $\widetilde{\nabla} _{\alpha'(t)} \alpha'(t) = \alpha''(t)$? If so, why is this true? – cos_dm_math21 Jan 22 '21 at 17:22
  • Yes, $h(X,Y)$ is the normal projection of $\widetilde{\nabla}XY$. And $\widetilde{\nabla}{\alpha'}\alpha'=\alpha''$ because covariant differentiation on $\Bbb R^{n+1}$ is just taking directional derivatives. – Ivo Terek Jan 22 '21 at 17:26
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    Got it! Thanks a lot! – cos_dm_math21 Jan 22 '21 at 17:55